Category: Engineering Mathematics Objective Questions

250+ TOP MCQs on Special Functions – 2 (Beta) and Answers

Ordinary Differential Equations Question Paper focuses on “Special Functions -2 (Beta)”.

1. β(m, n) = β(n, m). Is the statement true?
a) True
b) False
Answer: a
Explanation: L.H.S. = (beta(m, n) = frac{Gamma(m).Gamma(n)}{Gamma(m+n)}. )
R.H.S. ( = beta(n, m) = frac{Gamma(n).Gamma(m)}{Gamma(n+m)} = frac{Gamma(m).Gamma(n)}{Gamma(m+n)} = ) L.H.S.
Therefore, (beta(m, n) = beta(n, m).)

2. Which of the following function is not called the Euler’s integral of the first kind?
a) (beta(m, n) = int_0^1 x^{m-1} (1-x)^{n-1} dx (m>0, n>0) )
b) (beta(m, n) = int_0^{π/2} (sin⁡θ)^{2m-1} (cos⁡θ)^{2n-1} dθ )
c) (beta(m, n) = int_0^∞ frac{y^{n+1}}{(1+y)^{m+n}} dy )
d) (beta(m, n) = 2 int_0^{π/2} (sinθ)^{2m-1} (cos⁡θ)^{2n-1} dθ )
Answer: b
Explanation: Euler’s integral of the first kind is nothing but Beta function. So, here only ( beta(m, n) = int_0^{π/2}(sin⁡θ)^{2m-1} (cos⁡θ)^{2n-1} dθ) is not the definition of Beta function.

3. Which of the following is not the definition of Beta function?
a) (beta(m, n) = 2int_0^1 x^{m-1} (1-x)^{n-1} dx (m>0, n>0) )
b) (beta(m, n) = 2int_0^{π/2}(sin⁡θ)^{2m-1} (cos⁡θ)^{2n-1} dθ )
c) (beta(m, n) = int_0^∞ frac{y^{n+1}}{(1+y)^{m+n}} dy )
d) (beta(m, n) = int_0^1 frac{x^{m-1}+x^{n-1}}{(1+x)^{m+n}} dx )
Answer: a
Explanation: (beta(m, n)) can be written as either (int_0^1 x^{m-1} (1-x)^{n-1} dx , (m>0, n>0))
(or)
(= 2 int_0^{π/2}(sin⁡θ)^{2m-1} (cos⁡θ)^{2n-1} dθ )
(or)
( = int_0^∞ frac{y^{n+1}}{left(1+yright)^{m+n}} dy ;or int_0^1 frac{x^{m-1}+x^{n-1}}{left(1+xright)^{m+n}} dx).
So the correct answer is (2int_0^1 x^{m-1} (1-x)^{n-1} dx (m>0, n>0) ) which is actually not the formula for Beta function.

4. What is the value of (beta(m, frac{1}{2}) )?
a) β(m, m)
b) 22m-1 β(m, m)
c) 22m+1 β(m, m)
d) 22m β(m, m)
Answer: b
Explanation: (beta(m, frac{1}{2}) = 2int_0^{π/2}(sin⁡θ)^{2m-1} dθ )
(beta(m, m) = 2 int_0^{π/2}(sin⁡θ)^{2m-1} (cos⁡θ)^{2m-1} dθ )
( = 2^{-2m+2} int_0^{π/2}(2 sinθ cos⁡θ)^{2m-1} dθ )
Substituting 2θ=φ,
( = 2^{-2m+1} int_0^π sin⁡φdφ )
( = 2^{-2m+1}.2.int_0^{π/2}sin⁡φdφ )
( = frac{1}{2^{2m-1}} beta(m, frac{1}{2}). )

5. What is the value of β(3,2)?
a) (frac{1}{14} )
b) (frac{1}{16} )
c) (frac{1}{12} )
d) (frac{1}{10} )
Answer: c
Explanation: (beta(3, 2) = frac{Gamma(3).Gamma(2)}{Gamma(3+2)} )
( = frac{2!1!}{4!} = frac{1}{12}.)

6. What is the value of (beta(frac{1}{4},frac{3}{4}))?
a) (pi )
b) (sqrt{2}pi )
c) (sqrt{2pi} )
d) ({2}pi )
Answer: b
Explanation: (beta(frac{1}{4}, frac{3}{4}) = frac{Gamma(frac{1}{4}).Gamma(frac{3}{4})}{Gamma(frac{1}{4}+frac{3}{4})} )
( = frac{pi}{sin⁡ frac{π}{4}} = sqrt{2}pi. )

7. What is the value of (int_0^{π/2}sqrt{sin⁡θdθ} + int_0^{π/2}sqrt{cos⁡θ⁡dθ} )?
a) (8sqrt{pi} frac{Gamma(frac{3}{4})}{Gamma(frac{1}{4})} )
b) (4sqrt{pi} frac{Gamma(frac{3}{4})}{Gamma(frac{1}{4})} )
c) (8sqrt{pi} frac{Gamma(frac{1}{4})}{Gamma(frac{3}{4})} )
d) (4sqrt{pi} frac{Gamma(frac{1}{4})}{Gamma(frac{3}{4})} )
Answer: a
Explanation: (int_0^{π/2}sqrt{sin⁡θdθ} + int_0^{π/2}sqrt{cos⁡θ⁡dθ} )
( = beta(frac{3}{4}, frac{1}{2}) + beta(frac{3}{4}, frac{1}{2}) )
( = 2 beta(frac{3}{4}, frac{1}{2}) )
( = 2frac{Gamma(frac{1}{2}).Gamma(frac{3}{4})}{Gamma(frac{1}{2}+frac{3}{4})} )
( = 2sqrt{pi} frac{Gamma(frac{3}{4})}{Gamma(frac{1}{4})} . frac{1}{(frac{1}{4})} )
( = 8sqrt{pi} frac{Gamma(frac{3}{4})}{Gamma(frac{1}{4})}. )

8. What is the value of (int_0^∞ frac{1}{(1+y)^5} dy )?
a) 1⁄2
b) 1⁄3
c) 1⁄4
d) 1⁄5
Answer: c
Explanation: (int_0^∞ frac{y^{1-1}}{(1+y)^5} dy )
n=1 and m + n = 5 which implies m=4.
(beta(4, 1) = frac{Gamma(4).Gamma(1)}{Gamma(4+1)} )
(frac{3!0!}{4!} = frac{1}{4} ).

9. What is the value of (int_0^1 frac{dx}{sqrt{1+x^4}})?
a) (beta(frac{1}{4}, frac{1}{2}) )
b) (frac{1}{4sqrt{2}}beta(frac{1}{4}, frac{1}{2}) )
c) (frac{1}{3sqrt{2}}beta(frac{1}{3}, frac{1}{2}) )
d) (frac{1}{4sqrt{3}}beta(frac{1}{4}, frac{1}{3}) )
Answer: b
Explanation: Substitute x2 = tan⁡θ
Therefore θ varies from 0 to (frac{pi}{4}.)
(int_0^{π/4} frac{(secθ)^2}{2 sec⁡θ sqrt{tan⁡θ}} dθ )
(= int_0^{π/4}frac{dθ}{2sqrt{sin⁡θ cos⁡θ}} )
(= frac{1}{4sqrt{2}} beta(frac{1}{4}, frac{1}{2}).)

10. What is the value of (int_0^1frac{(y^5+y^2)}{(1+y)^9} dy )?
a) (frac{1}{158} )
b) (frac{2}{167} )
c) (frac{1}{146} )
d) (frac{1}{168} )
Answer: d
Explanation: m-1 = 5 => m=6 and n-1 = 2 => n=3.
(beta(6, 3) = frac{Gamma(6).Gamma(3)}{Gamma(6+3)} )
(= frac{5!2!}{8!} = frac{1}{168}. )

11. Solve using the Beta function. (int_0^1 x^{-2} (1-x)^{-3} dx. )
a) Can be solved using a Beta function with m = -1 and n = -2
b) Can be solved using a Beta function with m = 1 and n = -2
c) Can be solved using a Beta function with m = -1 and n = 2
d) Can’t be solved using the Beta function
Answer: d
Explanation: This function can’t be solved using the Beta function as m and n have negative values. Beta function can’t be solved if m and n are negative numbers.

12. What is the value of (int_0^∞ (sech x)^5 dx )?
a) (frac{3pi}{80} )
b) (frac{3pi}{240} )
c) (frac{3pi}{16} )
d) (frac{pi}{240} )
Answer: c
Explanation: (int_0^∞ (sech x)^5 dx = frac{2^5}{4} beta(frac{5}{2}, frac{5}{2}) )
(displaystyle = frac{8 Gamma(frac{5}{2})Gamma(frac{5}{2})}{Gamma(5)} = frac{(8*frac{3}{2}*frac{1}{2}*sqrt{pi}*frac{3}{2}*frac{1}{2}*sqrt{pi})}{24} = frac{9π}{48}=frac{3π}{16}. )

13. What is the value of (beta(frac{9}{2},3) )?
a) (frac{16}{1287} )
b) (frac{16}{1278} )
c) (frac{14}{1287} )
d) (frac{16}{127} )
Answer: a
Explanation: (beta(frac{9}{2},3) = frac{Gamma(frac{9}{2})Gamma(3)}{Gamma(frac{9}{2}+3)} )
( = frac{Gamma(frac{9}{2})2}{frac{13}{2}*frac{11}{2}*frac{9}{2}*Gamma(frac{9}{2})} = frac{16}{1287}.)

14. What is the value of (int_0^1 x^5 (1-x)^6 dx)?
a) (frac{1}{12*11*10*9*8*7} )
b) (frac{1}{12*11*10*9*8} )
c) (frac{1}{12*11*10*9*8*7*6} )
d) (frac{1}{12*11*10*9*8*7*6*5} )
Answer: c
Explanation: Here, m-1 = 5 which implies m=6 and n-1 = 6 which implies n=7.
(beta(6,7)= frac{Gamma(6)Gamma(7)}{Gamma(13)} = frac{1}{12*11*10*9*8*7*6}. )

Global Education & Learning Series – Ordinary Differential Equations.

To practice all questions papers on Ordinary Differential Equations,

250+ TOP MCQs on System of Equations and their Consistencies and Answers

Linear Algebra Interview Questions and Answers for freshers focuses on “System of Equations and their Consistencies”.

1. Test for consistency and solve to find the value of x.

5x + 3y + 7z = 4
3x + 26y + 2z = 9
7x + 2y + 10z = 5

a) Consistent, x=1
b) Consistent, x=-1
c) Inconsistent system, solution does not exist
d) Consistent, infinite number of solutions possible
Answer: d
Explanation: In this Question we have,
(begin{bmatrix}5&3&7\3&26&2\7&2&10end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}4\9\5end{bmatrix})
By 7R1 and 5R3
(begin{bmatrix}35&21&49\3&26&2\35&10&50end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}28\9\25end{bmatrix})
By R3-R1 and 5R2
(begin{bmatrix}35&21&49\15&130&10\0&-11&1end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}28\45\-3end{bmatrix})
By (R_2 – frac{3}{7} R_1)
(begin{bmatrix}35&21&49\0&121&-11\0&-11&1end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}28\33\-3end{bmatrix})
By (R_3 + frac{1}{11}R_2)
(begin{bmatrix}35&21&49\0&121&-11\0&0&0end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}28\33\0end{bmatrix})
By (frac{1}{7} R_1 and frac{1}{11} R_2)
(begin{bmatrix}5&3&7\0&11&-1\0&0&0end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}4\3\0end{bmatrix})
The rank of both the coefficient matrix and the augmented matrix is equal (i.e. 2)
Thus the equations are consistent.
However the rank of the matrix 2 is less than the total number of unknowns 3.
Hence the given set of equations has infinite number of solutions.

2. Test for consistency and solve the system of equations if possible to get the value of z.

2x - 3y+ 7z = 5
3x + y - 3z = 13
2x + 19y - 47z = 32

a) Consistent, z = -1
b) Consistent, z = 0
c) Inconsistent
d) Consistent, z = 5
Answer: c
Explanation: In this Question we have,
(begin{bmatrix}2&-3&7\3&1&-3\2&19&-47end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}5\13\32end{bmatrix})
By R1-R2
(begin{bmatrix}-1&-4&10\3&1&-3\2&19&-47end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}-8\13\32end{bmatrix})
By -R1
(begin{bmatrix}1&4&-10\3&1&-3\2&19&-47end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}8\13\32end{bmatrix})
By R2-3R1 and R3-2R1
(begin{bmatrix}1&4&-10\0&-11&27\0&11&-27end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}8\-11\16end{bmatrix})
By R3+R2
(begin{bmatrix}1&4&-10\0&-11&27\0&0&0end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}8\-11\5end{bmatrix})
Here we see that the rank of the co-efficient matrix is 2, while the rank of the augmented matrix is 3.
Since the two ranks are not equal, the given system of equations is inconsistent.

Global Education & Learning Series – Linear Algebra.

To practice all areas of Linear Algebra for Interviews,

250+ TOP MCQs on Surface Integrals and Answers

Linear Algebra Multiple Choice Questions on “Surface Integrals”.

1. Evaluate ∫xy dxdy over the positive quadrant of the circle x2+y2=a2.
a) (frac{a^4}{8} )
b) (frac{a^4}{4} )
c) (frac{a^2}{8} )
d) (frac{a^2}{4} )
Answer: a
Explanation: In the positive quadrant of the circle,
(y: 0 rightarrow a )
(x: o rightarrow sqrt{a^2-x^2} )
Therefore the integral is
(displaystyleint_0^a int_0^{sqrt{a^2-x^2}}xydxdy )
( = int_0^a frac{yx^2}{2} dy ) (from 0 to (sqrt{a^2-x^2}) )
( = frac{1}{2} int_0^a y(a^2-y^2) dy= frac{a^4}{8}. )

2. Evaluate ∫∫xy dxdy over the region bounded by x axis, ordinate x=2a and the curve x2=4ay.
a) (frac{a^4}{3} )
b) (frac{a^4}{6} )
c) (frac{a^3}{3} )
d) (frac{a^2}{6} )
Answer: a
Explanation: Both the curves meet at (2a,a).
Therefore,
(x:0 rightarrow 2a )
(y: 0 rightarrow frac{x^2}{4a} )
(int_0^2 a int_0^{frac{x^2}{4}}a xydxdy )
(= int_0^2 a frac{xy^2}{2} dx ) (from 0 to ( frac{x^2}{4a}) )
( = frac{1}{2} int_0^{2a} frac{x^5}{(16a^2 )} dx )
( = frac{a^4}{3}. )

3. Evaluate ∫∫x2+y2 dxdy in the positive quadrant for which x+y<=1.
a) (frac{1}{2} )
b) (frac{1}{3} )
c) (frac{1}{6} )
d) (frac{1}{12} )
Answer: c
Explanation: In this
x: 0 to 1
y:0 to 1-x
(int_0^1 int_0^{1-x} x^2+y^2 dxdy )
(= int_0^1 x^2 y+ frac{y^3}{3} dx ) (from 0 to 1-x)
(= int_0^1 x^2 (1-x)+ frac{(1-x)^3}{3} dx )
(= frac{1}{6}. )

4. Evaluate (int_0^∞ int_0^{π/2} e^{-r^{2}} rdθdr ).
a) ( pi )
b) ( frac{pi}{2} )
c) ( frac{pi}{4} )
d) ( frac{pi}{8} )
Answer: c
Explanation: The integral is in polar coordinates.
Substitute r2 as t
(int_0^∞ int_0^{π/2} e^{-t} dθ frac{dt}{2} )
(= frac{1}{2} int_0^{π/2}Γ(1)dθ )
(= frac{pi}{4}. )

5. Evaluate ∫∫rsinθdrdθ over the cardiod r = a(1+cosθ) above the initial line.
a) (4 frac{a^2}{3} )
b) ( frac{a^2}{3} )
c) (8 frac{a^2}{3} )
d) (4 frac{a^2}{6} )
Answer: a
Explanation: θ: 0 to π
r: 0 to a(1+cosθ)
(int_0^π int_0^{a(1+cosθ)} rsinθdrdθ )
(= int_0^π frac{r^2}{2} sinθdθ ) (from 0 to a(1+cosθ))
(= int_0^π frac{a^2}{2} (1+cosθ)^2 dθ )
(= 4 frac{a^2}{3}. )

6. Evaluate (int_0^∞ int_0^∞ e^{-(x^2+y^2 )} dxdy ) by changing into polar coordinates.
a) ( pi )
b) ( frac{pi}{2} )
c) ( frac{pi}{4} )
d) ( frac{pi}{8} )
Answer: c
Explanation: ( int_0^∞ int_0^∞ e^{-(x^2+y^2 )} dxdy )
(= int_0^{π/2} int_0^∞ e^{-(r^2 )} drdθ )
Substitute (r^2) as t
(= frac{1}{2} int_0^{π/2} int_0^∞ e^{-t} dtdθ )
( = frac{1}{2} int_0^{π/2}Γ(1)dθ )
( = frac{pi}{4}. )

7. Evaluate the following integral by transforming into polar coordinates.
(displaystyleint_0^a int_0^sqrt{a^2-x^2} ysqrt{x^2-y^2} dxdy )
a) ( frac{a^4}{2} )
b) ( frac{a^4}{3} )
c) ( frac{a^4}{4} )
d) ( frac{a^4}{5} )
Answer: c
Explanation: Subtitute x as rcosθ and y as rsinθ.
Therfore θ : 0 to Π/2
and r : 0 to a
(int_0^a int_0^{π/2} rsinθrrdrdθ )
(= [int_0^a r^3 dr][int_0^{π/2}sinθdθ] )
(= frac{a^4}{4} ).

8. Evaluate (int_0^∞ int_x^∞ frac{e^{-y}}{y} dydx ) by changing the order of integration.
a) 0
b) 1
c) 2
d) 1/2
Answer: b
Explanation: In the question, y: x to infnity
x: 0 to infinity
Now changing the orrder of integration:
y=x
y tends to infinity
y: 0 to infinity
x: 0 to y
(int_0^∞ int_0^y frac{e^{-y}}{y} dydx )
( = int_0^∞ frac{e^{-y}}{y} ydy )
= -(0-1)
= 1.

9. Calculate the area enclosed by parabolas x2 = y and y2 = x.
a) ( frac{1}{2} )
b) ( frac{1}{3} )
c) ( frac{1}{4} )
d) ( frac{1}{6} )
Answer: b
Explanation: x: 0 to 1
(y: x^2 , to , x^{1/2} )
(int_0^1 int_{x^2}^{sqrt{x}}dydx )
(= int_0^1 sqrt{x}-x^2 dx )
(= frac{2}{3} – frac{1}{3} )
(= frac{1}{3}. )

10. What is the area of a cardiod y = a(1+cosθ).
a) (frac{3πa^2}{2} )
b) (3πa^2 )
c) (frac{3πa^2}{4} )
d) (frac{3πa^2}{8} )
Answer: a
Explanation: θ : 0 to π
r : 0 to a(1+cosθ)
(frac{Area}{2} = int_0^π int_0^{a(1+cosθ)}rdrdθ )
(= frac{3πa^2}{4} )
Total area = ( 2* frac{3πa^2}{4} )
(frac{3πa^2}{2} ).

Global Education & Learning Series – Linear Algebra.

To practice all areas of Linear Algebra,

250+ TOP MCQs on Linear Difference Equations and Z – Transforms and Answers

Fourier Analysis Interview Questions and Answers focuses on “Linear Difference Equations and Z – Transforms”.

1. Find the Z-Transform of (^nC_p).
a) (1-z-1)n
b) (1+z-1)n
c) (1-z-1)-n
d) (1+z-1)-n
View Answer

Answer: b
Explanation: Using the Z-Transform formula, it can be written as
(Z(^nC_p) = 1+ ^nC_1 z^{-1}+ ^nC_2 z^{-2} + ^nC_3 z^{-3} )+……… which can be further equated to (1+z-1)n.

2. Find the function whose Z – Transform is (frac{1}{z} ).
a) δ(n)
b) δ(n+1)
c) U(n)
d) U(n+1)
View Answer

Answer: b
Explanation: δ(n) exists only at n=0 and δ(n+1) exists only at n=1. Therefore while substituting this function, the Z – Transform at every other place becomes zero except at n=1. Therefore the Z-Transform of δ(n+1) is (frac{1}{z}. )

3. Find the function whose Z transform is (e^{frac{1}{z}}).
a) log(n)
b) (frac{1}{n} )
c) (frac{1}{n!} )
d) (frac{1}{(n+1)!} )
View Answer

Answer: c
Explanation: Using the definition of Z- Transform we have (∑_{n=0}^∞ (frac{1}{n!} z^{-n})). Now, expanding this we get ( 1+frac{z^{-1}}{1} + frac{z^{-2}}{2} + frac{z^{-3}}{3} )+ …………. This is nothing but the expansion of (e^{frac{1}{z}}), hence the answer is (frac{1}{n!}.)

4. Find the inverse Z- Transform of ((frac{z}{z-a})^3).
a) (frac{1}{2} . (n+1) (n-2) a^{n-2} U(n) )
b) (frac{1}{2} . (n-1) (n-2) a^{n-3} U(n) )
c) (frac{1}{2} . (n-1) (n+2) a^{n-1} U(n) )
d) (frac{1}{2} . (n+1) (n+2) a^n U(n) )
View Answer

Answer: d
Explanation: The inverse Z-Transform of (frac{z}{z-a} ) is (a^n). The inverse Z-Transform of ((frac{z}{z-a})^2) is the convolution of an and an. Now, the inverse Z-Transform of ((frac{z}{z-a})^3) is the convolution of the result of the previous step with an an. Thus we get the answer (frac{1}{2} . (n+1) (n+2) a^n U(n) )

5. Find the inverse Z – Transform of (log frac{z}{z+1}).
a) (frac{(-1)^n}{n} )
b) (frac{(-1)^{n+1}}{n} )
c) (frac{1}{n} )
d) (frac{(-1)^n}{n+1} )
View Answer

Answer: a
Explanation: First, substitute z as ((frac{1}{y})) and then expand the got result. This is in the format of the Z-Transform expansion. Thus we get the required results.

6. Find the Z – Transform of sinh ⁡nθ.
a) (frac{sinh⁡θ}{z^2-2z cosh⁡θ+1} )
b) (frac{1}{2} frac{sinh⁡θ}{z^2-2z cosh⁡θ+1} )
c) (frac{z sinh⁡θ)}{z^2-2z cosh⁡θ+1} )
d) (frac{z(z-sinh⁡θ)}{z^2-2z cosh⁡θ+1} )
View Answer

Answer: a
Explanation: The first step to solve this is to expand the function sinh⁡nθ. The expansion of this function is of the form an. First we have to find the Z-Transform of 1 and then we have to use damping rule. To, get the answer, we take L.C.M.

7. Find the value of u3 if (U(z) = frac{3z^2+2z+10}{(z-1)^4} ).
a) 12
b) 13
c) 14
d) 15
View Answer

Answer: c
Explanation: Taking limz→∞ U(z), we get 0 which is u0. Now using the shifting property and again using the limit we get u1 which is 0. Again, by using the shifting property we get u2 which is 3. Now, by using shifting by 3 properties, we get the value of u3 which is 14.

8. Find the Z – Transform of np.
a) (-zfrac{d}{dz}(Z(n^{p-1})) )
b) (zfrac{d}{dz}(Z(n^p)) )
c) (-zfrac{d}{dz}(Z(n^{p+1})) )
d) (zfrac{d}{dz}(Z(n^{p+1})) )
View Answer

Answer: a
Explanation: The Z Transform of 1 can be found just by infinite G.P. sum. The Z- Transform can be found by differentiating the Z-Transform of 1 and multiplying by (-z). And the Z-Transform of n2, can also be found by differentiating the Z-Transform of n and multiplying by (-z). Hence the general form is (-zfrac{d}{dz}(Z(n^{p-1})) ).

9. The Z – Transform of a function is given by (U(z) = frac{z^3+6z^2+9z+3}{(z-1)^4}). Find the Z-Transform of un+2.
a) (frac{10z^3+3z^2+7z^1-1}{(z-1)^4} )
b) (frac{10z^4+3z^3+7z^2-z}{(z-1)^4} )
c) (frac{10z^4+4z^3+7z^2-2z}{(z-1)^4} )
d) (frac{10z^4+3z^3-4z}{(z-1)^4} )
View Answer

Answer: b
Explanation: First we have to find u0, which can be found by applying limits to U(z). Now shifting by 1 and then applying limits we get u1. Now using the second shift property, we find the Z-Transform of un+2.

10. Find u2 if (U(z) = frac{z^3+6z^2+9z+3}{(z-1)^4}).
a) 8
b) 9
c) 10
d) 11
View Answer

Answer: c
Explanation: The first step is to find the limit of the U(z), hence getting the u0. And again doing this we get u1. And again doing the shifting property, we get un+2. And doing the limits, we get the u2.

11. Find the order of the difference equation Δ3yn – Δ2yn – Δyn = 3.
a) 3
b) 4
c) 2
d) 5
View Answer

Answer: a
Explanation: The first step is to expand the given equation by replacing every Δyn by (yn+1– yn). Order of a difference equation is given by, (frac{n+3-n}{1}) which is actually 3.

12. Find the order of the difference equation yn+3 -3 yn+1 – yn-2 = 4.
a) 3
b) 4
c) 5
d) 6
View Answer

Answer: c
Explanation: The order of the given difference equation can be written as Order = (frac{n+3-n+2}{1}). Therefore the order is 5.

13. Find the difference equation of yn = A 3n + B 5n.
a) yn+2 -16 yn+1 + 15 yn-1 = 0
b) yn+3 -14 yn+1 + 30 yn = 0
c) 2 yn+2 -14 yn+1 + 15 yn = 0
d) 2 yn+2 -16 yn+1 + 30 yn = 0
View Answer

Answer: d
Explanation: This can be solved using the determinant.
(begin{bmatrix}
y_n & 1 & 1\
y_{n+1} & 3 & 5\
y_{n+2} & 9 & 25\
end{bmatrix} ) = 0. Now, by solving the determinant, we get the required difference equation.

14. Find the difference equation of y = ax + b.
a) Δ2y = 0
b) Δ2y = 1
c) Δ2y + 3Δy = 2
d) Δ2y + 4Δy = 5
View Answer

Answer: a
Explanation: First step is to take Δ operator on both sides of the given equation. Now, since here we have 2 unknown variables, we have take the Δ operator twice on both the sides, hence getting the required results.

15. Solve un+2 + 10 un+1 + 9 un = 2n.
a) (u_n = frac{2^{n+1}}{33}+frac{(-9)^{n+1}}{88}+frac{(-1)^{n+1}}{24} )
b) (u_n = frac{2^n}{33}+frac{(-9)^n}{88}+frac{(-1)^{n-1}}{24} )
c) (u_n = frac{2^{n+1}}{11}+frac{(-9)^{n+1}}{88}+frac{(-1)^n}{24} )
d) (u_n = frac{2^n}{11}+frac{(-9)^n}{88}+frac{(-1)^{n-1}}{24} )
View Answer

Answer: b
Explanation: Take Z – Transformation on both sides. Now keep U(z) on one side and take everything else to other side. Now by partial fractions method, we get it in the format of an. Now, taking inverse Z-Transform on both sides, we get the required results.

Global Education & Learning Series – Fourier Analysis.

To practice all areas of Fourier Analysis for Interviews, here is complete set of 1000+ Multiple Choice Questions and Answers.

 

250+ TOP MCQs on Taylor Mclaurin Series and Answers

Engineering Mathematics Multiple Choice Questions on “Taylor Mclaurin Series – 3”.

1. Expansion of function f(x) is?
a) f(0) + x⁄1! f‘ (0) + x2⁄2! f” (0)…….+xn⁄n! fn (0)
b) 1 + x⁄1! f‘ (0) + x2⁄2! f” (0)…….+xn⁄n! fn (0)
c) f(0) – x⁄1! f‘ (0) + x2⁄2! f” (0)…….+(-1)^n xn⁄n! fn (0)
d) f(1) + x⁄1! f‘ (1) + x2⁄2! f” (1)…….+xn⁄n! fn (1)
Answer: a
Explanation: By Maclaurin’s series, f(0) + x⁄1! f‘ (0) + x2⁄2! f” (0)…….+xn⁄n! fn (0)

2. The necessary condition for the maclaurin expansion to be true for function f(x) is __________
a) f(x) should be continuous
b) f(x) should be differentiable
c) f(x) should exists at every point
d) f(x) should be continuous and differentiable
Answer: d
Explanation: By Maclaurin’s series, f(0) + x⁄1! f‘ (0) + x2⁄2! f” (0)…….+xn⁄n! fn (0)

Where, f(x) should be continuous and differentiable upto nth derivative.

3. The expansion of f(a+h) is ______
a) (f(a)+frac{h}{1!} f'(a)+frac{h^2}{2!} f”(a)…….+frac{h^n}{n!} f^n (a))
b) (f(a)+frac{h}{1!} f'(a)+frac{h^2}{2!} f”(a)…….)
c) (hf(a)+frac{h^2}{1!} f'(a)+frac{h^3}{2!} f”(a)…….+frac{h^n}{n!} f^n (a))
d) (hf(a)+frac{h^2}{1!} f'(a)+frac{h^3}{2!} f”(a)………)
Answer: a
Explanation: By taylor expansion,
f(a+h) = f(a) + h⁄1! f'(a) + h2⁄2! f” (a)….

4. The expansion of eSin(x) is?
a) 1 + x + x2⁄2 + x4⁄8 +….
b) 1 + x + x2⁄2 – x4⁄8 +….
c) 1 + x – x2⁄2 + x4⁄8 +….
d) 1 + x + x3⁄6 – x5⁄10 +….
Answer: b
Explanation: Now f(x) = eSin(x), f(0) = 1
Hence, f‘ (x)=f(x)Cos(x), f‘ (0) = 1
f” (x)=f‘ (x)Cos(x) – f(x)Sin(x),f” (0)=1
f”’ (x)=f” (x)Cos(x) – 2f‘ (x)Sin(x) – f(x) cos⁡(x),f”’ (x) = 0
f”” (x)=f”’ (x)Cos(x) – 3f” (x)Sin(x) – 3f‘ (x) cos⁡(x) + f(x) sin⁡(x), f”” (x) = -3
Hence,
f(x) = eSin(x) = 1 + x + x2⁄2 – x4⁄8 +…. (By mclaurin’ sexpansion)

5. Expansion of y = Sin-1(x) is?
a) (x+frac{x^3}{6}+frac{3}{40} x^5+frac{5}{112} x^7+…..)
b) (x-frac{x^3}{6}+frac{3}{40} x^5-frac{5}{112} x^7+…..)
c) (frac{x^3}{6}-frac{3}{40} x^5+frac{5}{112} x^7…..)
d) (x+frac{x^2}{6}+frac{3}{40} x^2+frac{5}{112} x^2+…..)
Answer: a
Explanation: Given, y = Sin-1(x), hence at x = 0, y = 0
Now, differentiating it, we get
(frac{dy}{dx}=frac{1}{sqrt{1-x^2}}=(1-x^2)^{-1/2})
On expanding the R.H.S. by Binomial Theorem we get,
(frac{dy}{dx}=1+frac{x^2}{2}+frac{3}{8} x^4+frac{5}{16} x^6+…)
On integrating we get,
(y=x+frac{x^3}{6}+frac{3}{40} x^5+frac{5}{112} x^7+….+C)
By putting x=0 hence we get,
y=c=0
Hence,
(y=x+frac{x^3}{6}+frac{3}{40} x^5+frac{5}{112} x^7+….)

6. Find the expansion of f(x) = ln⁡(1+ex)?
a) (ln(2)+x/2+x^2/8-x^4/192+….)
b) (ln⁡(2)+x/2+x^2/8+x^4/192+….)
c) (ln⁡(2)+x/2+x^3/8-x^5/192+….)
d) (ln⁡(2)+x/2+x^3/8+x^5/192+….)
Answer: a
Explanation: Given, f(x) = ln⁡(1+ex), f(0) = ln⁡(2)
Differentiating it we get
(f{‘}(x)=frac{e^{x}}{1+e^x}=1-1/(1+e^x), f{‘}(0)=1/2)
Again differentiating we get
(f”(x)=e^x/(1+e^x)^2 =(f'(x))/(1+e^x),f”(0)=1/4)
(f”'(0)=((1+e^x) f”(x)-f'(x) e^x)/(1+e^x)^2 =(f”(x))/(1+e^x)-f'(x) f”(x)), hence f”'(0)=0
(f””(0)=(1(1+e^x) f”'(x)-f”(x) e^x)/(1+e^x)^2 -(f”(x))^2-f'(x) f”'(x)), hence f””(0)=1/8
Hence, by mclaurin’s series,
(f(x)=ln⁡(1+e^x)=ln⁡(2)+x/2+x^2/8-x^4/192+…)

7. Find the expansion of exSin(x)?
a) (e^{xSin(x)}=1+x^2-x^4/3+x^6/120-…)
b) (e^{xSin(x)}=1+x^2+x^4/3+x^6/120+…)
c) (e^{xSin(x)}=x+x^3/3+x^5/120+..)
d) (e^{xSin(x)}=x+x^3/3-x^5/120+…)
Answer: b
Explanation: Given, f(x) = exSin(x), f(0) = 1
Now, the expansion of xSin(x) is (x^2-x^3/3!+x^6/5!+…)
Hence, (e^xSin(x)=e^y=1+y+y^2/2!+y^3/3!+…)
Hence,
(e^{xSin(x)}=1+(x^2-frac{x^4}{3!}+frac{x^6}{6!}+..)+frac{(x^2-frac{x^4}{3!}+x^6/6!+..)^2}{2!}+frac{(x^2-frac{x^4}{3!}+x^6/6!+…)^3}{6}+..)
(e^{(xSin(x))}=1+x^2-frac{x^4}{3!}+frac{x^6}{5!}+frac{x^4}{2}-frac{x^6}{6}+frac{x^6}{6}+….) (we neglect all other other terms by considering the options given)
Hence, (e^{xSin(x)}=1+x^2+frac{x^4}{3}+frac{x^6}{120}+…)

8. Given f(x)= ln⁡(cos⁡(x)),calculate the value of ln⁡(cos⁡(π⁄2)).
a) -1.741
b) 1.741
c) 1.563
d) -1.563
Answer: a
Explanation: Given f(x) = ln⁡(Cos(x)), f(0) = 0
Differentiating it f'(x) = – tan⁡(x), f'(0) = 0
Again (f^{”}(x)=-sec^2⁡(x),f^{”}(0)=-1)
And (f^{”’}(x)=-2 sec⁡(x) ,sec⁡(x) ,tan⁡(x)=-2f^{”}(x)f'(x)), hence f”(0)=0
f””(x)=-2(f”'(x) f'(x)+(f”(x))2), hence f””(0)=-2
Now, by mclaurins’s series
f(x)=ln⁡(Cos(x))=(0+0-x^2/2!+0-x^4/12+..)
Therefore, f(x)=(-x^2/2!-x^4/12+…)
Hence,
ln⁡(cos⁡(π/2))=-1.741

9. Find the expansion of cos(xsin(t)).
a) (sum_{n=1}^∞ (frac{x^n [Cos(nt)]}{n!}))
b) (sum_{n=0}^∞ (frac{x^n [Cos(nt)]}{n!}))
c) (sum_{n=1}^∞ (frac{x^n [Sin(nt)]}{n!}))
d) (sum_{n=0}^∞ (frac{x^n [Sin(nt)]}{n!}))
Answer: b
Explanation:
Given, f(x)=Cos(xSin(t))=real part of (eixSin(t))
=real part of(exCos(t) eixSin(t))
=real part of(ex[Cos(t)+iSin(t)])
=Real part of (sum_{n=0}^∞ frac{x^n [Cos(nt)+iSin(nt)]}{n!})
=(sum_{n=0}^∞ frac{x^n [Cos(nt)]}{n!})

10. Find the expansion of Sin(lSin-1 (x)).
a) (lx-frac{l(1-l^2)}{3!}x^3+frac{l(1-l^2)(9-l^2)}{5!} x^5-…)
b) (lx+frac{l(1-l^2)}{3!} x^3+frac{l(1-l^2)(9-l^2)}{5!} x^5+…)
c) (1-lx^2+frac{l(1-l^2)}{3!} x^4-frac{l(1-l^2)(9-l^2)}{5!} x^6+…)
d) (1+lx^2+frac{l(1-l^2)}{3!} x^4+frac{l(1-l^2)(9-l^2)}{5!} x^6+…)
Answer: b
Explanation: Given, y = f(x) = Sin(lSin-1(x))
Now, differentiating,
(frac{dy}{dx}=Cos(lSin^{-1} (x))(frac{l}{sqrt{1-x^2}}))
Hence,
((1-x^2)(frac{dy}{dx})^2=lCos(lSin^{-1}(x))^2)
((1-x^2)(y_1)^2=l^2 Cos(lSin^{-1}(x))^2=l^2 [1-y^2])
Hence, differentiating again we get,
((1-x^2)2y_1 y_2-2xy_1^2=-2l^2 yy_1)
((1-x^2) y_2-xy_1+l^2 y=0)
Hence by Leibniz theorem,
((1-x^2) y_{(n+2)}-(2n+1)xy_{(n+1)}-(n^2-m^2) y_n=0)
Therefore by putting x=0, we get,
(y_{(n+2)}(0)=(n^2-l^2)y_n (0))
putting ,n=1,2,3,4,…..
(y_3 (0)=(1-l^2) y_1 (0)=l(1-l^2))
(y_4 (0)=(4-l^2) y_2 (0)=0)
(y_5 (0)=(9-l^2)3(0)=l(1-l^2)(9-l^2))
Hence,(y=Sin(lSin^{-1}(x))=lx+frac{l(1-l^2)}{3!}x^3+frac{l(1-l^2)(9-l^2)}{5!} x^5+…).

11. Expand (1 + x)1⁄x, gives ___________
a) e[1 + x⁄2 + 11x2⁄24 -…..]
b) e[1 – x⁄2 + 11x2⁄24 -…..]
c) e[x⁄2 – 11x2⁄24 -…..]
d) e[x⁄2 + 11x2⁄24 -…..]
Answer: b
Explanation: Given, y = (1 + x)1⁄x
Hence,
ln⁡(y)=(frac{ln⁡(1+x)}{x})
Hence, ln⁡(y)=(frac{1}{x} [x-frac{x^2}{2}+frac{x^3}{3}-……]=1-frac{x}{2}+frac{x^2}{3})-……
Hence, (y=e^{1-frac{x}{2}+frac{x^2}{3}-……}=e^{1+z}), where, (z=frac{-x}{2}+frac{x^2}{3}-……)
Hence, (y=e.e^z=e(1+z+frac{z^2}{2!}+frac{z^3}{3!}+…))
y=(e[1+(frac{-x}{2}+frac{x^2}{3}-……)+frac{1}{2!}(frac{-x}{2}+frac{x^2}{3}-…)^2+…])
y=(e[1-x/2+x^2/e+x^2/8+…])
y = e[1 – x⁄2 + 11x2⁄24 -…..].

12. Find the solution of differential equation, dy⁄dx = xy + x2, if y = 1 at x = 0.
a) (1-frac{x^2}{2!}+frac{3x^4}{4!}-frac{15x^6}{6!}+…)
b) (frac{x}{1!}+frac{3x^3}{4!}+frac{15x^5}{6!}+…)
c) (frac{x}{1!}-frac{3x^3}{4!}+frac{15x^5}{6!}-…)
d) (1+frac{x^2}{2!}+frac{3x^4}{4!}+frac{15x^6}{6!}+..)
Answer: d
Explanation: Given dy⁄dx = xy + x2
hence, dy⁄dy (x=0) = 0
and, d2y⁄dx2= xy1 + y + 2x
hence, y2 = xy1 + y + 2x
hence, d2y⁄dx2(x=0)=1
Differentiating it n times we get,
(y_{n+2}=xy_{n+1}+ny_n+y_n=xy_{n+1}+(n+1)y_n)
Putting x=0 we get,
(y_{n+2} (0)=(n+1)y_n (0))
Now putting the values of n as 1, 2, 3, 4, 5 we get,
(y_3 (0)=0, ,y_4 (0)=3, ,y_5 (0)=0, ,y_6(0)=15……) and so on
By mclaurin’s series,
(y=1+frac{x^2}{2!}+frac{3x^4}{4!}+frac{15x^6}{6!}+…)

250+ TOP MCQs on Variable Treated as Constant and Answers

Differential and Integral Calculus Multiple Choice Questions on “Variable Treated as Constant”.

1. If z=3xy+4x2, what is the value of (frac{∂z}{∂x})?
a) 3y+8x
b) 3x+4x2
c) 3xy+8x
d) 3y+3x+8x
View Answer

Answer: a
Explanation: Given: z=3xy+4x2
Using partial differentiation, we need to differentiate the function z with respect to x keeping y as constant. Thus, (frac{∂z}{∂x}=3y+8x.)

2. The value of (frac{∂z}{∂y})=8x2+6xy2+4. What is the function z expressed as?
a) z=8x3+2x2 y2+4x
b) z=8x2 y+2xy3+4y
c) z=8y+2xy2+4y
d) z=16x+6y2
View Answer

Answer: b
Explanation: Since the given partial differentiation is with respect to y, to find the expression for the function z, we need to integrate the given partial differentiation expression with respect to y.
Therefore, (∫frac{∂z}{∂y}.dy)=∫(8x2+6xy2+4).dy
z=8x2 y+(frac{6}{3}) xy3+4y
z=8x2 y+2xy3+4y

3. Find the correct values for (frac{∂f}{∂x} ,and, frac{∂f}{∂y}) for the function (f=frac{2}{x^3}y^2+4y^3.)
a) (frac{∂f}{∂x}= frac{-6}{x^2}, frac{∂f}{∂y}= frac{2}{x^3} y+8y^2)
b) (frac{∂f}{∂x}= frac{2}{x^4}, frac{∂f}{∂y}= frac{2}{x^3} y+12y^2)
c) (frac{∂f}{∂x}= frac{-6}{x^4}, frac{∂f}{∂y}= frac{4}{x^3} y+12y^2)
d) (frac{∂f}{∂x}= frac{-6}{x^4}, frac{∂f}{∂y}= frac{4}{x^3} y^2+12)
View Answer

Answer: c
Explanation: Given: (f=frac{2}{x^3}y^2+4y^3)
Differentiating partially with respect to x we get,
(frac{∂f}{∂x}=2frac{∂(x^{-3})}{∂x}=2(-3x^{-3-1}) = frac{-6}{x^4})
Differentiating with respect to y we get,
(frac{∂f}{∂y}= frac{4}{x^3}y+12y^2)

4. Volume of an object expressed in spherical coordinates is given by (V = int_0^{frac{π}{2}}∫_0^{frac{π}{3}}∫_0^1r cos∅ dr d∅ dθ.) The value of the integral is _______
a) (frac{sqrt{3}}{2})
b) (frac{1}{√2} π)
c) (frac{sqrt{3}}{2} π)
d) (frac{sqrt{3}}{8}π)
View Answer

Answer: d
Explanation: Given: (V = int_0^{frac{π}{2}}∫_0^{frac{π}{3}}∫_0^1r cos∅ ,dr ,d∅ ,dθ)
(V = int_0^{frac{π}{2}}∫_0^{frac{π}{3}}(frac{r^2}{2})_0^1cos∅ ,d∅ ,dθ)
(V = frac{1}{2} int_0^{frac{π}{2}}(sin∅)_0^{frac{π}{3}}d∅ ,dθ)
(V = frac{1}{2}×frac{sqrt{3}}{2}int_0^{frac{π}{2}}dθ)
(V = frac{1}{2}×frac{sqrt{3}}{2}×frac{π}{2})
(V = frac{sqrt{3}}{8}π )

5. Which of the following relations hold true for division rule of differentiation?
a) ((frac{f(x)}{g(x)})’= frac{f'(x)}{g'(x)} )
b) ((frac{f(x)}{g(x)})’= frac{g(x) f'(x)- g'(x)f(x)}{(f(x))^2})
c) ((frac{f(x)}{g(x)})’= frac{g(x)f'(x)- g'(x)f(x)}{(g(x))^2} )
d) ((frac{f(x)}{g(x)})’= frac{f(x)g'(x)-f'(x)g(x)}{(g(x))^2} )
View Answer

Answer: c
Explanation: The division rule of differentiation for two functions is given by,
((frac{f(x)}{g(x)})’= frac{g(x)f'(x)- g'(x)f(x)}{(g(x))^2} )

6. What is the value of (frac{∂^2z}{∂x∂y}) for the z=3x2y+5y?
a) 3xy
b) 6x
c) 3x+5
d) 6xy
View Answer

Answer: b
Explanation: Given: z=3x2y+5y
(frac{∂}{∂x}(frac{∂z}{∂y})=frac{∂}{∂x}(3x^2+5)=6x )

7. Which of the following is correct?
a) (frac{d}{dx} (sin^{-1}(⁡x))= frac{1}{sqrt{1-x^2}})
b) (frac{d}{dx} (sec^{-1}(⁡x))= frac{1}{sqrt{x^2-1}})
c) (frac{d}{dx} (tan^{-1}(⁡x))= frac{1}{sqrt{x^2+1}})
d) (frac{d}{dx} (sin^{-1}(⁡x))= frac{1}{x+1} )
View Answer

Answer: a
Explanation: Rules for derivatives of inverse trigonometric functions are:

  • (frac{d}{dx} (sin^{-1}⁡(⁡x))= frac{1}{sqrt{1-x^2}})
  • (frac{d}{dx} (sec^{-1}⁡(⁡x))= frac{1}{xsqrt{x^2-1}})
  • (frac{d}{dx} (tan^{-1}(⁡x))= frac{1}{1+x^2})

8. Given (∫_0^8x^{frac{1}{3}}dx,) find the error in approximating the integral using Simpson’s 1/3 Rule with n=4.
a) 1.8
b) 2.9
c) 0.3
d) 0.35
View Answer

Answer: d
Explanation: Given: (∫_0^8x^{frac{1}{3}}dx, n = 4,)
Let (f(x)= x^{frac{1}{3}},)
(∆x = frac{b-a}{2}=frac{8-0}{2}=4) ………………since b=8, a=0 (limits of the given integral)
Hence endpoints xi have coordinates {0, 2, 4, 6, 8}.
Calculating the function values at xi, we get,
(f(0)= 0^{frac{1}{3}}=0)
(f(2)= 2^{frac{1}{3}})
(f(4)= 4^{frac{1}{3}})
(f(6)= 6^{frac{1}{3}})
(f(8)= 8^{frac{1}{3}} =2)
Substituting these values in the formula,
(∫_0^8x^{frac{1}{3}}dx ≈ frac{∆x}{3} [f(0)+4f(2)+2f(4)+4f(6)+f(8)])
( ≈ frac{2}{3} [0+4(2^{frac{1}{3}})+2(4^{frac{1}{3}})+ 4(6^{frac{1}{3}})+2] ≈ 11.65)
Actual integral value,
( ∫_0^8x^{frac{1}{3}}dx= left(frac{x^{frac{4}{3}}}{frac{4}{3}}right)_0^8=12)
Error in approximating the integral = 12 – 11.65 = 0.35

9. The determinant of the matrix whose eigen values are 6, 4, 3 is given by ___________
a) 3
b) 24
c) 72
d) 13
View Answer

Answer: b
Explanation: The product of the eigen values of a matrix gives the determinant of the matrix,
Therefore, ∆ = 72.

10. The symbol used for partial derivatives, ∂, was first used in mathematics by Marquis de Condorcet.
a) True
b) False
View Answer

Answer: a
Explanation: Partial derivatives are indicated by the symbol ∂. This was first used in mathematics by Marquis de Condorcet who used it for partial differences.

Global Education & Learning Series – Differential and Integral Calculus.

To practice all areas of Differential and Integral Calculus, here is complete set of 1000+ Multiple Choice Questions and Answers.